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The protons alpha to the oxygen in methyl propionate are more shielded than their corresponding protons in ethyl propionate.

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We can see from the NMR data that this is the case as well.

My TA says this is due to the inductively withdrawing effect of the extra carbon in ethyl propionate.

However, this strikes me as blatantly wrong. The carbon should be inductively donating ... I don't see how it pulls electron density away from the specified hydrogens.

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In your example we are comparing the electron releasing/withdrawing properties of a hydrogen substituent and a methyl substituent, and the effect this has on proton ($\ce{H_{a}}$) chemical shifts.

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Carbon (2.55) is more electronegative than hydrogen (2.2). Therefore, the methyl substituent will be more electron withdrawing compared to the hydrogen substituent. This means that the methyl group will remove electron density from the carbon to which it is attached, compared to the hydrogen substituent. The fact that this carbon has lower electron density in the methyl case, is further transmitted to the protons ($\ce{H_{a}}$) attached to this carbon. Consequently the $\ce{H_{a}}$ protons will also experience lower electron density in the R=methyl case.

Lower electron density at an nmr-active nucleus results in deshielding (downfield shift). Consequently, the $\ce{H_{a}}$ protons are deshielded in the ethyl ester (R=methyl) relative to the methyl ester (R=hydrogen), in agreement with the spectral data you've provided.

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    $\begingroup$ Are you saying that the electron withdrawing effects of carbon act through another carbon atom (as is apparently the case in the ethyl ester)? $\endgroup$ – Dissenter Jan 30 '15 at 4:11
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    $\begingroup$ Yes, we're comparing the effect of hydrogen vs. methyl through a carbon atom and on thehydrogens attached to that carbon atom. $\endgroup$ – ron Jan 30 '15 at 4:16
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    $\begingroup$ After all, the standard shifts for methyl, methanediyl and methanetriyl groups are 0.9, 1.3, and 1.7 respectively. $\endgroup$ – RBW Jan 30 '15 at 14:48

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