An organic compound gave the following spectroscopic data.Deduce the structure:UV : 220 (ε 1800) nmIR : 1745, 1608, 1497, 1456c * m ^ - 1'H NMR in delta / 1.25 * (s, J = 7Hz, 3H) 2(t, 3H)(q, J = 7Hz, 2H )
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UV: The absorption at 220 nm suggests the presence of a carbonyl group (C=O) or an alkene (C=C) in the compound. Show more…
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Compound $\mathbf{B}$ is isomeric with $\mathbf{A}$ (Problem 19.66 ) and shows an IR peak at $1715 \mathrm{cm}^{-1} .$ The $^{1} \mathrm{H}$ NMR spectrum of B has peaks at $2.4 \delta(1 \mathrm{H},$ septet, $J=7 \mathrm{Hz}), 2.1 \delta(3 \mathrm{H}, \text { singlet }),$ and $1.2 \delta(6 \mathrm{H}, \text { doublet, } J=7 \mathrm{Hz}) .$ What is the structure of B?
Give the structure of the compound with molecular mass $=88$ and the following spectra. Proton NMR: $\delta 1.2(6 H, \mathrm{~d}, J=7 \mathrm{~Hz}) ; \delta 2.5(1 H,$ septet, $J=7 \mathrm{~Hz}) ; \delta 10(1 H,$ broad s $)$ IR: $\quad 2600-3400 \mathrm{~cm}^{-1}$ (broad) $, 1720 \mathrm{~cm}^{-1}$
Three isomeric chloro-derivatives of pyridine (A, B and $\mathbf{C}$ ) analyse as containing $40.58 \%$ C. $2.04 \% \mathrm{H}$ and $9.46 \%$ N. The $^{1}$ H NMR spectroscopic data for the compounds are as follows where d = doublet, $\mathrm{dd}=$ doublet of doublets and $\mathrm{t}=$ triplet: $$\begin{array}{cl} \text { Compound } & ^{1} \text { H NMR } \delta / \text { ppm } \\ \text { A } & 7.66(\mathrm{t}, J=7.6 \mathrm{Hz}) \\ & 7.31(\mathrm{d}, J=7.6 \mathrm{Hz}) \\ \text { B } & 8.64(\mathrm{d}, J=2.1 \mathrm{Hz}) \\ & 8.25(\mathrm{t}, J=2.1 \mathrm{Hz}) \\ \text { C } & 8.70(\mathrm{dd}, J=3.0 \text { and } 0.3 \mathrm{Hz}) \\ & 8.13(\mathrm{dd}, J=9.0 \text { and } 3.0 \mathrm{Hz}) \\ & 7.68(\mathrm{dd}, J=9.0 \text { and } 0.3 \mathrm{Hz}) \\ \hline \end{array}$$ In each isomer, Cl atoms are in either the 2 - or 3 position with respect to the N atom. Suggest structures for $\mathbf{A}, \mathbf{B}$ and $\mathbf{C}$.
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