Hi Briana!

It's based on an arbitrary list of functional group priorities set by the International Union of Pure and Applied Chemistry (IUPAC).

When there's a tie between a double bond and a triple bond for the purpose of determining the lowest locant, the double bond will always take priority.

-Ben

Hi Selamawit!

This image has important information that's cropped out.  Can you provide an image that contains the whole thing?

-Ben

Hi Erika!

See the attached image.

The "hex" prefix means we have six carbons in the parent chain.

The "3,3-dichloro" means we have two chlorine atoms bonded to carbon 3 on the six-atom chain.

The "5-en" means we have a carbon-carbon double bound between carbons 5 and 6.

The "2-ol" means we have a hydroxyl group at carbon 2.

The "(R)" means that carbon 2, a chiral center, has an R configuration.  We have lessons on stereochemistry that go into detail about what this means.  First we assign Cahn-Ingold-Prelog priorities to each group bonded to the chiral center.  These are shown in green.  Next, we imagine the lowest-priority group (the H) in the back.  If the group 1 > group 2 > group 3 sequence follows the CLOCKWISE direction, then it's R.  If the group 1 > group 2 > group 3 sequence follows the COUNTERCLOCKWISE direction, then it's S.  In this case, since it's CLOCKWISE, it's R.

-Ben  

Hi Valentina!

Inversion/retention applies when your start and end with a chiral center:

     Inversion: configuration of chiral center changes

     Retention: configuration of chiral center does not change

For the top two products, which show the -OH taking the place of the -Br, yes, we can use inversion/retention to describe these two.  The top left molecule would be the product of retention, and the top right molecule would be the product of inversion.

We would NOT use inversion/retention to describe the formation of the bottom two products, since the carbon bearing the -OH group was not a chiral center in the starting material.

In order for a molecule to have an enantiomer, it must be non-superimposable on its mirror image.  Usually, this means the molecule has at least one chiral center (four different groups bonded to a carbon).  Whether it's a starting material or a product makes no difference.  We look at each molecule individually.

-Ben

Hi Nicole!

See the attached image.There are two SN2 reactions back-to-back, each of which takes place with inversion.

For the first reaction, with PBr3, we can represent the inversion by switching the places of the hydrogen and methyl group on carbon 1 (methyl group goes from wedged to dashed).

For the second reaction, with KCN, we can represent the inversion, again, by switching the places of the hydrogen and methyl group on carbon 1 (methyl group goes from dashed back to wedged).

-Ben

Hi Yessica!

Yes.

The =C-H bonds are IR active and would typically show up as a peak with medium intensity around 3020 - 3100 cm^-1.

The C-H bonds in the methyl group at the top right of the molecule are also IR active and would typically show up as a peak with medium intensity around 2850 - 2960 cm^-1.

-Ben

Hi Mona!

To be clear, the reason why we get an M+2 peak in the mass spectrum of a compound that contains chlorine is because chlorine has two naturally occurring isotopes, chlorine-35 and chlorine-37, which have relative abundances of 75.77% and 24.23%, respectively.

Since the M+2 peak is due to chlorine isotopes, not carbon isotopes, we cannot use the M+2 peak in the calculation to determine the number of carbons in the molecule.

-Ben 

Hi Mona!

If you're asking if the methyl group should be in front of the plane of the screen (wedged) and the hydrogen atom should be behind the plane of the screen (dashed), the answer is no.

See the attached image.  I've drawn it a little bit differently to better show the nature of SN2 substitution.  Imagine the bonds to the hydrogen atom, the methyl group, and the CPhEtO- carbon form an umbrella-type shape.  Upon SN2 substitution, the umbrella turns inside out, and the three bonds that were pointing downward are now pointing upward, but the wedges and dashed lines haven't changed.

This can be difficult to see if the molecule is a ring.  I've provided another example, SN2 substitution of 1-fluoro-1-iodoethane to help get the point across. 

Hi Mona!

Yes, it's entirely possible for the second step of the mechanism to be SN1 substitution at the carbon to the left of the oxygen with SN2 substitution at the carbon to the left of the oxygen happening later.

If the reaction took place in a polar aprotic solvent, that would favor SN2 happening first.  If a protic solvent was used, then that would favor SN1.  Since the solvent isn't shown, we don't know for sure, so we can assume that both mechanisms would take place in parallel.

-Ben

Hi Valentina!

The four groups attached to the bromine-bearing carbons are the following:

A bromine atom

A hydrogen atom

A -CH2-CH2-CHBr-CH2-CH2-*back to original carbon to complete the ring*

A -CH2-CH2-CHBr-CH2-CH2-*back to original carbon to complete the ring*

Since those last two groups are equivalent, and since a chiral center must be bonded to four NON-equivalent groups, the bromine-bearing carbons are not chiral centers.

It's not a meso compound.  Meso compounds are compounds that contain chiral centers but are not chiral molecules due to a plane of symmetry in the molecule.

-Ben