See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. The compound shown is 2-bromo-1,5-hexadiene: CH2=C(Br)-CH2-CH2-CH=CH2. It contains two double bonds and already one bromine atom. Step 2: Determine the reaction. Two equivalents of Br2 undergo anti addition (electrophilic addition) across each of the two double bonds in CCl4 solvent. Each double bond reacts with one Br2 molecule via anti addition, adding two Br atoms across each double bond. Step 3: Identify the new stereocenters created. - Addition of Br2 to the terminal double bond CH=CH2 (the right end) creates two new stereocenters: the two carbons of that former double bond each bear different substituents after addition. - Addition of Br2 to the CH2=C(Br)- end: the C(Br) carbon already has a Br and after addition becomes C(Br)(Br)- which has two bromines — this carbon is NOT a stereocenter (two identical Br groups on the same carbon means it cannot be a stereocenter). The terminal CH2 becomes CHBr- which also is not a stereocenter since the carbon at the end becomes -CH2Br (two H's on terminal carbon). Step 4: Re-examine the structure carefully. The starting material from the image is: CH2=C(Br)-CH2-CH=CH2 — a five-carbon compound (2-bromo-1,3-pentadiene? or 2-bromo-1,4-pentadiene?). From the image, there appear to be two CH2 groups between the double bonds: CH2=C(Br)-CH2-CH2-CH=CH2 (2-bromo-1,5-hexadiene). Step 5: Addition of Br2 to CH2=C(Br)- gives BrCH2-CBr2- (the central carbon now has 2 Br's, not a stereocenter; the terminal carbon is CH2Br, not a stereocenter). Step 6: Addition of Br2 to -CH=CH2 gives -CHBr-CH2Br. The internal carbon -CHBr- is a stereocenter (R or S). The terminal -CH2Br is not (two H's). So only ONE new stereocenter is formed from this addition. Step 7: With only one new stereocenter from the right-side double bond addition, we get 2 stereoisomers (R and S configuration at that center). The anti addition gives a specific relative configuration at the two carbons of the double bond — but since one of them (CH2Br) is not a stereocenter, only one stereocenter results, giving 2 stereoisomers total. Step 8: Therefore, 2 stereoisomeric pentabromides are formed. Options (b), (c), (d) are incorrect because there is only one true stereocenter generated in the product, yielding exactly 2 stereoisomers. Therefore, the correct answer is A.