Jump to content

Chem P1 doubts - Org Chem

Recommended Posts

Hello,

could somebody explain the attached problems? Also, for Q13, how can you infer the polarity with a rough idea of electronegativity? Do you need to know the values for some elementsCapture.PNG.25b71daaf0e5e3169d8967b2cb50692c.PNGCapture.PNG.f5b34a5f5f7920b7f291e9b0e1a98d45.PNG?

Thanks in advance!

Share this post


Link to post
Share on other sites

Without databooklets you just have to know 13. Rememeber that alkanes are non polar, but ketones, haloalkanes, and amines are all polar. 13. B

Electrophilic substitution adds something to the aromatic ring carbon, not on a substituent. 36. C

A lot of enantiomers contain a carbon bonded to 4 different groups. In 37. B, the third carbon from the left is bonded to an ethyl group, hydrogen, bromine, and a methyl group. 

  • Like 1

Share this post


Link to post
Share on other sites

13. 

The answer is C-H in the CH4

a. C=O : 3.4-2.6= .8 || b. C-H : 2.6-2.2= .4 || c. C-Cl : 3.2-2.6= .6 || d. N-H : 3.0-2.2= .8 (For reference; I am aware you don't get these values on Paper 1)

You can deduce based on the trend that C-H will be less polar than N-H and that C=O will be more polar than C-Cl.

So, now we are down to C-H and C-Cl.

We know that Cl has a relatively high electronegativity as it's on the far right. Which means that C-Cl has more polarity. (For the most part, if one of the options has a N, O, F, or Cl they are going to be the most polar option.)

So, C-H is the answer. 

  • Like 1

Share this post


Link to post
Share on other sites
4 hours ago, kw0573 said:

Without databooklets you just have to know 13. Rememeber that alkanes are non polar, but ketones, haloalkanes, and amines are all polar. 13. B

Electrophilic substitution adds something to the aromatic ring carbon, not on a substituent. 36. C

A lot of enantiomers contain a carbon bonded to 4 different groups. In 37. B, the third carbon from the left is bonded to an ethyl group, hydrogen, bromine, and a methyl group. 

Ohhh thank you!

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

×