Given below are two statements :
Statement I : Compound (X), shown below, dissolves in \( NaHCO_3 \) solution and has two chiral carbon atoms.
Statement II : Compound (Y), shown below, has two carbons with \( sp^3 \) hybridization, one carbon with \( sp^2 \) and one carbon with \( sp \) hybridization.
In the light of the above statements, choose the correct answer from the options given below :
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An alkyne (\( -C \equiv C- \)) always contains \textbf{two} \( sp \) hybridized carbons. An allene (\( C=C=C \)) contains \textbf{one} \( sp \) hybridized carbon (the middle one).
Step 1: Understanding the Concept:
Solubility in sodium bicarbonate indicates the presence of an acidic group (like -COOH). Chiral centers are carbons with four unique substituents.
Hybridization is determined by the number of sigma bonds and lone pairs (linear = sp, trigonal planar = \( sp^2 \), tetrahedral = \( sp^3 \)).
Step 3: Detailed Explanation: Statement I:
Compound (X) contains a carboxylic acid group (-COOH), which reacts with \( NaHCO_3 \) to liberate \( CO_2 \), hence it dissolves.
The structure \( -C^*H(OH)C^*H(CH_3)COOH \) has two chiral carbons (indicated by asterisks). Both are bonded to four different groups. Thus, Statement I is true. Statement II:
Analyzing \( CH_3-CH_2-C \equiv C-C(=O)H \) :
- The \( CH_3 \) and \( CH_2 \) carbons are \( sp^3 \) hybridized (4 sigma bonds). Count = 2.
- The carbonyl carbon (\( C=O \)) is \( sp^2 \) hybridized (3 sigma bonds, 0 lone pairs). Count = 1.
- The two carbons in the triple bond (\( C \equiv C \)) are both \( sp \) hybridized (2 sigma bonds each). Count = 2.
The statement says there is one carbon with \( sp \) hybridization. Since there are two, the statement is false.
Step 4: Final Answer:
Statement I is true, but statement II is false.
