See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution ## Step 1: Identify the starting material and first reaction Starting material: $CH_3C \equiv CH$ (propyne) Adding $HCl$ (2 moles) to an alkyne follows Markovnikov's rule. The first mole adds to give an intermediate vinyl chloride, and the second mole adds to the resulting double bond. ## Step 2: Addition of first HCl mole $$CH_3C \equiv CH + HCl \rightarrow CH_3CCl=CH_2$$ (1,1-dichloroalkene intermediate) ## Step 3: Addition of second HCl mole The second $HCl$ adds to the double bond following Markovnikov's rule (H adds to the carbon with more H atoms): $$CH_3CCl=CH_2 + HCl \rightarrow CH_3CCl_2CH_3$$ **Product A:** $CH_3CCl_2CH_3$ (1,1-dichloropropane) ## Step 4: Elimination with aqueous KOH and heat Aqueous $KOH$ with heat causes **elimination**, not substitution. With two chlorines on the same carbon (geminal dichloride), this undergoes **double elimination** to regenerate the triple bond: $$CH_3CCl_2CH_3 \xrightarrow{aq.KOH, \text{Heat}} CH_3C \equiv CH$$ **Product B:** $CH_3C \equiv CH$ (back to propyne) ## Conclusion The products are: - **A:** $CH_3CCl_2CH_3$ ✓ - **B:** $CH_3C \equiv CH$ ✓ **Answer: (C)**