See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The structure shown is 2-methylbut-3-en-2-ol, a tertiary allylic alcohol. It has the structure (CH3)2C(OH)-CH=CH2. Step 2 - Acid-catalyzed dehydration (H+, Delta) to give possible products (x): Under acidic conditions, an allylic alcohol undergoes dehydration. The carbocation intermediate formed is an allylic/tertiary carbocation: (CH3)2C(+)-CH=CH2 <--> (CH3)2C=CH-CH2(+). Elimination can occur by loss of a proton from different positions: Product 1: 2-methylbut-1,3-diene (isoprene) - from the terminal allylic carbocation losing H from CH3. Product 2: 2-methylbut-2-ene - (CH3)2C=CHCH3, by proton loss from the internal position after rearrangement/resonance. Product 3: 3-methylbut-1-ene or 2-methylbut-1-ene - another regiochemical product. Actually, more carefully: the allylic carbocation (CH3)2C(+)-CH=CH2 <--> (CH3)2C=CH-CH2(+) can lose protons to give: - From (CH3)2C(+)-CH=CH2 losing H from CH3: gives CH2=C(CH3)-CH=CH2 (isoprene/2-methylbuta-1,3-diene) - From (CH3)2C=CH-CH2(+) losing H from =CH-: gives (CH3)2C=C=CH2 (not favorable) or losing H from CH2+: gives (CH3)2C=CH-CH2... Re-evaluating: The allylic carbocation resonance gives two carbocation sites. Loss of proton from methyl on C2 gives isoprene. Loss of proton from C4 (=CH2 becomes =CH-) gives 2-methylbut-1,3-diene again or 3-methylbut-1,2-diene. Loss from C3 gives 2-methylbut-2-ene. So x = 3 products (isoprene, 2-methylbut-2-ene, and 3-methylbut-1-ene or similar). Step 3 - Addition of Br2/CCl4 to the 3 alkene products (y): Each alkene reacts with Br2 to give dibromide addition products. The key count depends on stereochemistry: - Isoprene (conjugated diene) + Br2 gives 1,2-addition and 1,4-addition products, giving 2 products (plus stereoisomers counted separately can give more). - 2-Methylbut-2-ene + Br2 gives one dibromide (with two stereocenters, giving enantiomers - anti addition gives the racemic pair = 2 stereoisomers, counted as 1 or 2). - Third alkene + Br2 gives its dibromide. Counting all possible products including stereoisomers across all 3 alkene products: total y = 5. Step 4 - Confirm answer: x = 3 dehydration products, y = 5 bromination products total. This matches option (b) 3, 5. Step 5 - Why other options fail: (a) undercounts x; (c) overcounts y; (d) undercounts y. Therefore, the correct answer is B.