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Which of the following compound has minimum C-H bond length?

  1. ethane
  2. ethene
  3. 1,2-dibromoethene
  4. 1,2-dibromoethane

How can we know the comparison of $\ce{C-H}$ bond length (not $\ce{C-C}$)? Also, there is not any reagent given, so that we would do dehydrohalogenation reaction at it. How to compare it simply?

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One way to compare $\ce{C-H}$ bond lengths is to compare their stretching frequencies using IR spectroscopy. However, since ethane and ethylene are gases at ambient conditions, we are going to have some difficulties in accuracy. Yet, it is worth comparing. Since we are comparing $\ce{C-H}$ bond in each case, we can suggest that the stronger the stretching frequency of $\ce{C-H}$ bond is, the shorter the bond should be.

1,2-Dibromoethane is a liquid at room temperature and $\pu{760 torr}$ with boiling point in the range of $\pu{131-132 ^\circ C}$. 1,2-Dibromoethene is from Aldrich is a mixture of cis- and trans-isomers, which boils at $\pu{110 ^\circ C}$ under $\pu{754 torr}$ pressure. IR spectra of both are extracted from NIST.gov webbook and illustrated as follows:

1,2-Dibromoethene: Range of $\ce{C-H}$ stretching is around $\pu{3150-3050 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3100 cm^{-1}}$):

1,2-dibromoetene

1,2-Dibromoethane: Range of $\ce{C-H}$ stretching is around $\pu{3010-2950 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3000 cm^{-1}}$):

1,2-dibromoetane

Accordingly, we can easily suggest that $\ce{C-H}$ bond in 1,2-dibromoethane is longer than that in 1,2-dibromoethene. However, this is hardly a surprise since we are comparing a $\mathrm{sp^3-s}$ bond versus a $\mathrm{sp^2-s}$ bond. Nonetheless, this is a good exercise because it confirm the validity of the test, in a way. Yet, the $\ce{C-H}$ stretching at around $\pu{3000 cm^{-1}}$ is undoubtedly suggested that $\ce{C-H}$ bond in 1,2-dibromoethane is shorter than a $\ce{C-H}$ bond in a regular alkane.

Now, let's see two good liquid alkane and alkene examples to represent gasious ethane and ethene, respectively.

2-Methylpent-1-ene (a terminal alkene) is a liquid at room temperature and $\pu{760 torr}$ with boiling point of $\pu{62 ^\circ C}$. n-Pentane (the shortest n-alkene at ambient condition) is a liquid at room temperature and $\pu{760 torr}$ with its boiling point around the range of $\pu{35-36 ^\circ C}$.

2-Methylpent-1-ene: Range of $\ce{C-H}$ stretching is around $\pu{3100-2800 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{3075 cm^{-1}}$):

2-Methylpent-1-ene

2-Methylpent-1-ene: Range of $\ce{C-H}$ stretching is around $\pu{2990-2800 cm^{-1}}$ (Maximum $\ce{C-H}$ stretching is around $\pu{2900 cm^{-1}}$):

enter image description here

Thus, we can confidently suggest that the $\ce{C-H}$ bond lengths of given compounds are in following order:

ethane > 1,2-dibromoethane >ethene > 1,2-dibromoethene

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    $\begingroup$ Unfortunately vibrational spectroscopy is not a technique with which to obtain structure. There is only a general trend in bond length of, say, single bonds, vs. frequency so any answers based on spectra are just guesses. Rotational spectroscopy in the gas phase can and is used to obtain bond length for small molecules. $\endgroup$ – porphyrin May 20 '18 at 7:42
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    $\begingroup$ Your comments are well taken. I'm giving a relative comparison to OP's request for simple explanation. Otherwise, we are all speculating a lot of things, which can't explain otherwise. $\endgroup$ – Mathew Mahindaratne May 20 '18 at 7:55
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    $\begingroup$ Got it, Thanks for the explanation sir :) !.. $\endgroup$ – Chirag Lathi May 20 '18 at 9:16
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According to me, answer should be (c) less than (b) less than (d) less than (a). Reason- Due to lone pairs present on bromine, in (c) the Carbon would acquire a partial sp character. But in (d), due to higher electronegativity of $\ce{Br}$ w.r.t $\ce{H}$, more p character would be in $\ce{C-Br}$ than $\ce{C-H}$ due to which $\ce{C-H}$ bond length would reduce.

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  • $\begingroup$ @Mathew Mahindaratne, is my ans correct ?? $\endgroup$ – Aakhyat Singh May 20 '18 at 3:41
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    $\begingroup$ Your argument is logical. :-) $\endgroup$ – Mathew Mahindaratne May 20 '18 at 3:44

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