Given that spectroscopic analysis of compound A with formula C 5 H 10 O yields the following data: IR: 3400cm -1 ; 1640 cm -1 1 HNMR: 1.63 δ (3H, singlet); 1.70 δ (3H, singlet); 3.83 δ (1H, broad singlet); 4.15 δ (2H, doublet, J =7Hz); 5.70 δ (1H,triplet, J=7Hz). i) How many double bonds and/or rings are there in A has is to be stated. ii) The identity of the oxygen containing functional group is to be derived from IR spectral data. iii) The protons responsible for the NMR data listed are to be identified. iv) A structure for the compound A is to be proposed. Concept introduction: In IR the O-H stretching of alcohols and phenols are observed in the range 3200-3550 cm -1 . The C=C stretching is observed in the range 1620-1680 cm -1 . In 1HNMR spectrum, the alcoholic proton absorption occurs in the range 3.4 δ - 4.5 δ while that of phenolic proton occur in the range 3.0 to 8.0. The aromatic protons give a broad peak in the range 6.5 δ - 8.0 δ, the benzylic protons normally absorb in the region 2.3 δ - 3.0 δ. The absorption due to 10 alkyl group (CH 3 ) is seen around 0.7 δ - 1.3 δ, that due to a 20 alkyl group (CH 2 ) is seen around 1.2 δ - 1.6 δ while that due to 30 alkyl group (CH) is seen around 1.4 δ - 1.8 δ The multiplicity of a signal gives an idea about the protons present in the adjacent carbons. To state: The number of the double bonds and/or rings in the compound A. To identity: The oxygen containing functional group from IR spectral data and the protons responsible for the NMR data listed. To propose: A structure for the compound A.

Question

A chemist synthesized compound X as a racemic mixture. When the ketone group in X was enzymatically reduced to the corresponding alcohol, a 100% yield was obtained of the product shown below. Choose the statement that best describes this result. ОН enzyme C;H1 `OCH,CH; pH 4.0 C3H1 `OCH,CH3 ОН ÕH X (racemic) (100% yield) One enantiomer of compound X reacts quickly with the enzyme. The other enantiomer of compound X is unreactive, but rapidly equilibrates with the reactive enantiomer under the reaction conditions. Since compound X was racemic, it makes sense that only a single product was obtained. O The product is a meso compound, so either enantiomer of compound X gives the same product. One enantiomer of compound X reacts quickly with the enzyme, while the other enantiomer of compound X remains unchanged.

Answer 1

Question

Chapter 17.SE, Problem 66AP

Interpretation Introduction

Interpretation:

Given that spectroscopic analysis of compound A with formula C₅H₁₀O yields the following data:

IR: 3400cm^-1; 1640 cm^-1

¹HNMR: 1.63 δ (3H, singlet); 1.70 δ (3H, singlet); 3.83 δ (1H, broad singlet); 4.15 δ (2H, doublet, J =7Hz); 5.70 δ (1H,triplet, J=7Hz).

i) How many double bonds and/or rings are there in A has is to be stated. ii) The identity of the oxygen containing functional group is to be derived from IR spectral data. iii) The protons responsible for the NMR data listed are to be identified. iv) A structure for the compound A is to be proposed.

Concept introduction:

In IR the O-H stretching of alcohols and phenols are observed in the range 3200-3550 cm^-1. The C=C stretching is observed in the range 1620-1680 cm^-1. In 1HNMR spectrum, the alcoholic proton absorption occurs in the range 3.4 δ - 4.5 δ while that of phenolic proton occur in the range 3.0 to 8.0. The aromatic protons give a broad peak in the range 6.5 δ - 8.0 δ, the benzylic protons normally absorb in the region 2.3 δ - 3.0 δ.

The absorption due to 10 alkyl group (CH₃) is seen around 0.7 δ - 1.3 δ, that due to a 20 alkyl group (CH₂) is seen around 1.2 δ - 1.6 δ while that due to 30 alkyl group (CH) is seen around 1.4 δ - 1.8 δ The multiplicity of a signal gives an idea about the protons present in the adjacent carbons.

To state:

The number of the double bonds and/or rings in the compound A.

To identity:

The oxygen containing functional group from IR spectral data and the protons responsible for the NMR data listed.

To propose:

A structure for the compound A.

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