Name the following compound:

CH3
I
CH = CH2 - CH3
I
CH2
I
CH - CH3
I
CH2
I
CH
I
CH3

2-ethyl-4-methylheptane
2-ethyl-4-methylheptene
3-methyl-5-propyl-2-hexene
3,5-dimethyl-2-octene

3,5-dimethyl-2-octene

Explanation:

Please note that there is no H at carbon 3 less carbon becomes penta hydra.

The compound is:

CH3

I

C = CH2 - CH3

I

CH2

I

CH - CH3

I

CH2

I

CH2

I

CH3

To name the above compound, do the following:

1. Locate the longest continuous chain i.e octene

2. Start counting from the side that gives the double bond the lowest low count since the double bond is the functional group. In doing this, the double bond is at carbon 2.

3. Locate the substituent groups attached and their position in the parent chain. In doing so, you will see that there are two CH3 group attached and they are at carbon 3 and carbon 5. Since the substituents attached are the same, we'll name them as 'dimethyl' indicate that they are two methyl groups

Now, we'll combine the above findings in order to obtain the name. Therefore, the name of the compound is:

3,5-dimethyl-2-octene

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2