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NCERT Exemplar · Q12

Q.Which of the following alkenes on ozonolysis give a mixture of ketones only? (Note: more than one of the given options may be correct.)

(i) CH3-CH=CH-CH3
(ii) CH3-C(CH3)=CH2
(iii) cyclopentylidene=C(CH3)2 (a cyclopentane ring joined by a double bond to C(CH3)2)
(iv) (CH3)2C=C(CH3)2
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Ozonolysis turns each doubly-bonded carbon into a carbonyl; a ketone results only when that carbon bears two alkyl groups (no H). Only (iii) and (iv) have both alkene carbons fully substituted, so they give ketones only.

Reductive ozonolysis (O3\text{O}_3, then Zn/H2O\text{Zn/H}_2\text{O}) cleaves a C=C\text{C=C} and caps each carbon with = ⁣O=\!\text{O}. A double-bond carbon carrying two alkyl groups becomes a ketone; one carrying an H becomes an aldehyde (or formaldehyde if it bears two H). For "ketones only," both alkene carbons must be fully alkyl-substituted.

(i) CH3-CH=CH-CH3\text{CH}_3\text{-CH=CH-CH}_3 (but-2-ene). Each alkene carbon has one methyl and one H:

CH3-CH=CH-CH3  → O3; Zn/H2O   2 CH3CHO\text{CH}_3\text{-CH=CH-CH}_3 \;\xrightarrow{\ \text{O}_3;\ \text{Zn/H}_2\text{O}\ }\; 2\,\text{CH}_3\text{CHO}

Two molecules of acetaldehyde — aldehydes, not ketones.

(ii) (CH3)2C=CH2\text{(CH}_3\text{)}_2\text{C=CH}_2 (2-methylpropene). The left carbon bears two methyls; the right carbon bears two H:

(CH3)2C=CH2  → O3; Zn/H2O   (CH3)2CO+HCHO\text{(CH}_3\text{)}_2\text{C=CH}_2 \;\xrightarrow{\ \text{O}_3;\ \text{Zn/H}_2\text{O}\ }\; \text{(CH}_3\text{)}_2\text{CO} + \text{HCHO}

Acetone plus formaldehyde — a ketone and an aldehyde, so not ketones only.

(iii) cyclopentylidene=C(CH3)2\text{C(CH}_3\text{)}_2. The ring carbon of the double bond is bonded to two ring CH2\text{CH}_2 groups (no H); the other carbon bears two methyls:

→ O3; Zn/H2O   cyclopentanone+(CH3)2CO\xrightarrow{\ \text{O}_3;\ \text{Zn/H}_2\text{O}\ }\; \text{cyclopentanone} + \text{(CH}_3\text{)}_2\text{CO}

Cyclopentanone and acetone — two ketones only. ✓ …

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