See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify constraints: (i) Molecular formula C7H14 => degree of unsaturation = (2*7+2-14)/2 = 1, so exactly one double bond (alkene). (ii) Compound (A) shows geometrical (cis/trans) isomerism => the double bond must have two different groups on each carbon of the C=C. (iii) Product (B) gives Cannizzaro reaction => (B) is an aldehyde with no alpha-hydrogen (since Cannizzaro requires an aldehyde lacking alpha-H). This means one fragment from ozonolysis is a formaldehyde (HCHO) or a benzaldehyde, but since we are dealing with aliphatic C7H14, the relevant case is that one carbonyl carbon has no alpha-H, i.e., it is of the form R3C-CHO where R groups leave no alpha-H, meaning the carbon adjacent to CHO bears no H — effectively a pivalaldehyde-type: (CH3)3C-CHO. Step 2 - Analyze option (c): (CH3)3C-CH=CH-CH3 This is 4,4-dimethyl-2-pentene: the double bond is between C2 and C3 (using a 5-carbon chain with two methyls at C4... let's count: (CH3)3C-CH=CH-CH3). Carbon count: (CH3)3C = 4 carbons, CH=CH = 2 carbons, CH3 = 1 carbon => total 7 carbons. Formula C7H14. Correct. Geometrical isomerism: The double bond carbons are CH= (bearing H and the (CH3)3C group) and =CH (bearing H and CH3). Both carbons have two different substituents => cis/trans isomers exist. Condition (ii) satisfied. Ozonolysis (O3 then Zn/AcOH, reductive workup): (CH3)3C-CH=CH-CH3 => (CH3)3C-CHO + CH3-CHO B = (CH3)3C-CHO (trimethylacetaldehyde / pivalaldehyde) and C = CH3CHO (acetaldehyde). Does (B) = (CH3)3C-CHO give Cannizzaro reaction? Pivalaldehyde has no alpha-hydrogen (the alpha carbon is the quaternary C(CH3)3 with no H). Yes, it undergoes Cannizzaro reaction. Condition (iii) satisfied. Step 3 - Check other options: (a) CH3-CH(CH3)-C(CH3)=CH-CH3: ozonolysis gives CH3-CH(CH3)-CO-CH3 (a ketone, not an aldehyde) + CH3CHO. Ketones do not give Cannizzaro. Fails. (b) (CH3)3C-CH2-CH=CH2: terminal alkene, the =CH2 carbon has two identical H substituents => no geometrical isomerism. Fails condition (ii). (d) (CH3)2C(CH3)-CH2-CH=CH2: same issue, terminal alkene, no geometrical isomerism. Fails condition (ii). Step 4 - Conclusion: Only option (c) satisfies all conditions: C7H14, geometrical isomerism, and ozonolysis product B = pivalaldehyde giving Cannizzaro reaction. Therefore, the correct answer is C.