Q.An alkylhalide with molecular formula C6H13Br on dehydro halogenation gave two isomeric alkenes X and Y with molecular formula C6H12. On reductive ozonolysis, X and Y gave four compounds CH3COCH3, CH3CHO, CH3CH2CHO and (CH3)2CHCHO. Find the alkylhalide.

Step 1. Pair up the four ozonolysis products by carbon count so that each pair sums to 6 carbons (matching the C6H12 alkenes X and Y): acetone (CH3COCH3, 3C) + propionaldehyde (CH3CH2CHO, 3C) sums to 6C, and acetaldehyde (CH3CHO, 2C) + isobutyraldehyde/(CH3)2CHCHO (4C) also sums to 6C -- these must be the two product-pairs from X and Y respectively.

Step 2. Reconstruct X from its pair: a carbon giving acetone on cleavage must have carried TWO methyl groups (no H) before cleavage, and a carbon giving propionaldehyde must have carried an ethyl group and one H -- joining these two alkene carbons back together gives X = (CH3)2C=CH-CH2-CH3, i.e. 2-methylpent-2-ene.

Step 3. Reconstruct Y from its pair: a carbon giving acetaldehyde must have carried a methyl group and one H, and a carbon giving isobutyraldehyde must have carried an isopropyl group and one H -- joining these gives Y = CH3-CH=CH-CH(CH3)-CH3, i.e. 4-methylpent-2-ene. …

Work BACKWARD from each ozonolysis product pair to its parent alkene (matching total carbon count and each fragment's own substitution pattern), then find the single alkyl halide w …

• Pairing the four carbonyl products incorrectly (e.g. acetone with acetaldehyde) without first checking that each pair's total carbon count equals 6 -- both alkenes X and Y must independently account for all 6 carbons of the parent C6H13Br. …