I'm having trouble understanding how to "estimate" the value of the BDE for the compound shown below (I believe it is a secondary benzylic C-Br value) shown below (the 58 kcal/mol). We are supposed to be able to estimate the value from the table values shown below. My teacher said to use the C-H values and C-Br values from the table below to estimate it, but I don't understand how they are able to indicate the value. Any helpful drawings would be nice since I'm new to organic chemistry and have trouble just understanding the names! Thank you.
What you have shown in red is the second propagation step in the benzylic free radical bromination of ethylbenzene. The bond dissociation energy (BDE) of bromine is +46 kcal/mol while the overall heat of reaction of the step is -12 kcal/mol. Thus, the formation of the C-Br bond must be -58 kcal/mol. The BDE for the C-Br bond in (1-bromoethyl)benzene is +58 kcal/mol.
However, the BDE for bromine is not listed in your tables. So how can one estimate the BDE of the C-Br bond in (1-bromoethyl)benzene using the data in the tables? A simple ratio and proportion will suffice. In equation A compare the primary C-H bond of ethane to that of toluene. This ratio is compared to the ratio of the BDE of the secondary C-H bond of propane and its counterpart in ethylbenzene, which is the unknown "x". An estimated value of 82 kcal/mol is obtained. In equation B, this ratio compares secondary C-H bonds with secondary C-Br bonds. The estimate for the BDE of the C-Br bond in (1-bromoethyl)benzene is the value of "y" which is equal to 59 kcal/mol. Pretty close, wouldn't you say?