A mixed ether (P), when heated with excess of hot concentrated hydrogen iodide produces two different alkyl iodides which when treated with aq. NaOH give compounds (Q) and (R). Both (Q) and (R) give yellow precipitate with NaOI. Identify the mixed ether (P):A schematic diagram showing the transformation sequence from ether (P) to products (Q) and (R) giving a yellow precipitate.
A.textOption 1$\text{Option 1}$
A schematic diagram showing the transformation sequence from ether (P) to products (Q) and (R) giving a yellow precipitate.
B.textOption 2$\text{Option 2}$
C.textOption 3$\text{Option 3}$
A schematic diagram showing the transformation sequence from ether (P) to products (Q) and (R) giving a yellow precipitate.
D.textOption 4$\text{Option 4}$
A schematic diagram showing the transformation sequence from ether (P) to products (Q) and (R) giving a yellow precipitate.
Solution & Explanation
### Core Logic
We work backwards. Both (Q) and (R) give a positive iodoform test (yellow precipitate with NaOI). Since (Q) and (R) are obtained by treating alkyl iodides with aq. NaOH, they must be alcohols.
Alcohols that give a positive iodoform test must possess the structural unit CH_3-CH(OH)-$CH_3-CH(OH)-$. Therefore, both alkyl iodides must contain the CH_3-CH(I)-$CH_3-CH(I)-$ group.
The precursor mixed ether (P) was treated with excess concentrated HI to yield these two alkyl iodides. For both sides of the ether to produce iodides capable of forming secondary alcohols with terminal methyls upon hydrolysis, the ether must be composed of two branches that look like -CH(CH_3)-$-CH(CH_3)-$.
Let's evaluate the correct option (1):
Structure (P) is an ether linking a sec-butyl group and a secondary-pentyl group (or similar structure). Excess HI cleaves the ether to give two secondary alkyl iodides.
Hydrolysis with aq. NaOH converts both alkyl iodides into secondary alcohols.
Both secondary alcohols have a methyl group directly adjacent to the hydroxyl-bearing carbon, fully satisfying the iodoform test requirement.
### Pattern Recognition
Excess hot HI cleaves ethers (R-O-R'$R-O-R'$) into two moles of alkyl iodides (R-I$R-I$ and R'-I$R'-I$). Only secondary alcohols with a methyl group or ethanol itself give positive iodoform tests. Look for an ether containing CH_3-CH(O-)-$CH_3-CH(O-)-$ branching on BOTH sides.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Class 12 Chemistry: Haloalkanes and Haloarenes
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
A schematic diagram showing the transformation sequence from ether (P) to products (Q) and (R) giving a yellow precipitate.
Keywords:#mixed ether (P)#JEE Main 2026 Evening Q66#Alcohols, Phenols and Ethers JEE Main 2026#Reactions of Ethers JEE Main 2026#Reaction flow#Ether cleavage#Hydrogen iodide#Iodoform test
More Alcohols, Phenols and Ethers Previous-Year Questions
Given below are two statements:
Statement I: Phenol on treatment with CHCl_3/aq$CHCl_{3}/aq$. KOH under refluxing condition, followed by acidification produces p-hydroxy benzaldehyde as the major product and o-hydroxy benzaldehyde as the minor product.
Statement II: The mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be easily separated through steam distillation.
In the light of the above statements, choose the correct answer from the options given below:
A.textBoth Statement I and Statement II are false$\text{Both Statement I and Statement II are false}$
B.textStatement I is true but Statement II is false$\text{Statement I is true but Statement II is false}$
C.textBoth Statement I and Statement II are true$\text{Both Statement I and Statement II are true}$
D.textStatement I is false but Statement II is true$\text{Statement I is false but Statement II is true}$
Solution
### Related Formula
Reimer-Tiemann Reaction: Phenol xrightarrowCHCl_3, KOH$\xrightarrow{CHCl_3, KOH}$ o-hydroxybenzaldehyde (Salicylaldehyde) as major product.
### Core Logic
Statement I claims that p-hydroxybenzaldehyde is the major product. This is incorrect. In the Reimer-Tiemann reaction, ortho-hydroxybenzaldehyde (salicylaldehyde) is the major product due to the stabilization of the intermediate and product via intramolecular hydrogen bonding.
Reimer-Tiemann reaction mechanism diagram
Statement II states that the mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be separated by steam distillation. This is true. Ortho-hydroxybenzaldehyde forms intramolecular hydrogen bonds, making it more volatile (steam volatile), whereas the para-isomer forms intermolecular hydrogen bonds, increasing its boiling point and making it non-steam volatile.
### Step 1: Final Conclusion
Therefore, Statement I is false but Statement II is true.
### Pattern Recognition
Always remember: Ortho-isomers capable of intramolecular H-bonding are steam volatile. Para-isomers exhibit intermolecular H-bonding and are not steam volatile.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Alcohols Phenols and Ethers
Q69jee_main_2026_22_january_eveningSynthesis of 3,3-Dimethyl-2-butanol
3,3-Dimethyl-2-butanol cannot be prepared by:
A. Displays five synthetic routes to evaluate preparation of 3,3-dimethyl-2-butanol.
B. Displays five synthetic routes to evaluate preparation of 3,3-dimethyl-2-butanol.
C. Displays five synthetic routes to evaluate preparation of 3,3-dimethyl-2-butanol.
D. Displays five synthetic routes to evaluate preparation of 3,3-dimethyl-2-butanol.
E. Displays five synthetic routes to evaluate preparation of 3,3-dimethyl-2-butanol.
Choose the correct answer from the options given below:
A. B only
B. B and E only
C. B and C only
D. B, C and E only
Solution
### Related Formula
textAcid-catalyzed hydration: Carbocation formation rightarrow textEthyl/Methyl Shift$$\text{Acid-catalyzed hydration: Carbocation formation } \rightarrow \text{Ethyl/Methyl Shift}$$textOxymercuration-Demercuration: Markovnikov addition without carbocation rearrangement$$\text{Oxymercuration-Demercuration: Markovnikov addition without carbocation rearrangement}$$
### Core Logic
Step 1: Evaluate Route B:
Acid catalyzed hydration of 3,3-dimethyl-1-butene involves carbocation formation followed by 1,2-methyl shift to give 2,3-dimethyl-2-butanol as major product instead of 3,3-dimethyl-2-butanol.
Step 2: Evaluate Route E:
Oxymercuration-demercuration or specific hydration route E fails to give the targeted alcohol structural framework.
Hence, 3,3-Dimethyl-2-butanol CANNOT be prepared by routes B and E.
Displays five synthetic routes to evaluate preparation of 3,3-dimethyl-2-butanol.Displays five synthetic routes to evaluate preparation of 3,3-dimethyl-2-butanol.Displays five synthetic routes to evaluate preparation of 3,3-dimethyl-2-butanol.Displays five synthetic routes to evaluate preparation of 3,3-dimethyl-2-butanol.
### Pattern Recognition
Sees: Carbocation rearrangement during acid hydration.
Shortcut: Acid-catalyzed hydration of (CH_3)_3C-CH=CH_2$(CH_3)_3C-CH=CH_2$ undergoes methyl shift yielding 2,3-dimethyl-2-butanol, failing to produce 3,3-dimethyl-2-butanol.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Class 11 Chemistry: Hydrocarbons
Q65jee_main_2026_24_january_morningAcetylation of Hydroxyl Groups
A hydroxy compound (X) with molar mass 122 text g mol^-1$122 \text{ g mol}^{-1}$ is acetylated with acetic anhydride, using a large excess of the reagent ensuring complete acetylation of all hydroxyl groups. The product obtained has a molar mass of 290 text g mol^-1$290 \text{ g mol}^{-1}$. The number of hydroxyl groups present in compound (X) is :
A.3$3$
B.5$5$
C.2$2$
D.4$4$
Solution
### Core Logic
During the acetylation of an alcohol, an -textH$-\text{H}$ atom (molar mass = 1) is replaced by an acetyl group (-textCOCH_3$-\text{COCH}_3$, molar mass = 43).
Therefore, the net increase in molar mass per hydroxyl group acetylated is:
Delta M = 43 - 1 = 42 text g/mol$$\Delta M = 43 - 1 = 42 \text{ g/mol}$$
Given:
Molar mass of starting compound (X) = 122 text g/mol$122 \text{ g/mol}$
Molar mass of completely acetylated product = 290 text g/mol$290 \text{ g/mol}$
Total increase in molar mass:
290 - 122 = 168 text g/mol$$290 - 122 = 168 \text{ g/mol}$$
### Step 1: Calculate Number of Hydroxyl GroupstextNumber of -textOH groups = fractextTotal Mass IncreasetextIncrease per group = frac16842 = 4$$\text{Number of } -\text{OH groups} = \frac{\text{Total Mass Increase}}{\text{Increase per group}} = \frac{168}{42} = 4$$Acetylation mechanism showing mass increaseAcetylation mechanism showing mass increase
### Pattern Recognition
Acetylation replaces H$H$ with COCH_3$COCH_3$. Always divide the mass difference by 42 to find the number of -OH$-OH$ (or -NH_2$-NH_2$) groups.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Class 12 Chemistry: Biomolecules
Q68jee_main_2026_24_january_morningReactions with Active Metals and Bicarbonate
Consider the following two reactions A and B.
(A) textPhenol xrightarrowtextNa textMain Product + textgas (x)$\text{Phenol} \xrightarrow{\text{Na}} \text{Main Product} + \text{gas (x)}$
(B) textBenzoic acid xrightarrowtextNaHCO_3 textMain Product + textgas (y)$\text{Benzoic acid} \xrightarrow{\text{NaHCO}_3} \text{Main Product} + \text{gas (y)}$
Numerical value of [molar mass of x+ molar mass of y] is
A.4$4$
B.88$88$
C.46$46$
D.160$160$
Solution
### Core Logic
Reaction (A): Phenol + Sodium metal
Phenol reacts with active metals like Sodium to form sodium phenoxide and evolve hydrogen gas.
2 C_6H_5OH + 2 Na rightarrow 2 C_6H_5O^-Na^+ + H_2 uparrow$$2 C_6H_5OH + 2 Na \rightarrow 2 C_6H_5O^-Na^+ + H_2 \uparrow$$
Gas (x) is H_2$H_2$. Molar mass of H_2 = 2 text g/mol$H_2 = 2 \text{ g/mol}$.
Reaction (B): Benzoic acid + Sodium bicarbonate
Benzoic acid is a sufficiently strong acid to react with weak bases like NaHCO_3$NaHCO_3$, undergoing decarboxylation to evolve carbon dioxide gas.
C_6H_5COOH + NaHCO_3 rightarrow C_6H_5COO^-Na^+ + H_2O + CO_2 uparrow$$C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COO^-Na^+ + H_2O + CO_2 \uparrow$$
Gas (y) is CO_2$CO_2$. Molar mass of CO_2 = 44 text g/mol$CO_2 = 44 \text{ g/mol}$.
### Step 1: Final Calculation
textSum of molar mass = M(H_2) + M(CO_2)$$\text{Sum of molar mass} = M(H_2) + M(CO_2)$$textSum = 2 + 44 = 46$$\text{Sum} = 2 + 44 = 46$$
### Pattern Recognition
Active metals + acids/alcohols = H_2$H_2$ gas. Bicarbonates + acids (stronger than carbonic acid) = CO_2$CO_2$ effervescence.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Alcohols, Phenols and Ethers
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
More Alcohols, Phenols and Ethers Questions — jee_main_2026_23_january_evening
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