draw the alkyne formed when 3,4-dichloroheptane is treated with an excess of a strong base such as sodium amide.
I got CH3CH3C(triple bond)CCH2CH3 but was told it was the wrong answer
To determine the correct answer, we need to analyze the reaction between 3,4-dichloroheptane and a strong base such as sodium amide. This reaction involves the elimination of two molecules of hydrogen chloride (HCl) from the starting compound.
Here's a step-by-step approach to solving this problem:
1. Start by identifying the structure of 3,4-dichloroheptane, which is a seven-carbon chain with chlorine substituents on the third and fourth carbons:
CH3CH2CH(Cl)CH(Cl)CH2CH2CH3
2. Next, determine the position of the strong base, sodium amide (NaNH2). Since it is an excess, it will abstract hydrogen atoms from different carbon atoms in the molecule.
3. The sodium amide will deprotonate the hydrogen atoms on the carbon adjacent to chlorine atoms, resulting in the formation of two alkenes. Remember that deprotonation by a strong base results in the formation of an anion.
CH3CH2C(Cl)CH(Cl)CH2CH2CH3
Deprotonation of the hydrogen atom adjacent to one of the chlorine atoms results in the formation of an alkene with a double bond between the second and third carbons. Similarly, deprotonation of the hydrogen atom adjacent to the other chlorine atom results in a second alkene with a double bond between the fifth and sixth carbons.
So, the correct answer should be:
CH3CH=CHCH(Cl)CH2CH2CH3 + CH3CH2CH=CHCH2CH2CH3
It appears that your initial answer was not correct, possibly due to an error in the position of the triple bond in relation to the chlorine substituents. Double-checking the position of the chlorine atoms before deprotonating the adjacent hydrogens should lead you to the correct answer.