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Chapter 5: Q5-49P (page 222)

Draw the skeletal structure of 3,3-dimethyl-7-(1-methyl ethyl)-6-(1-methyl propyl) decane.

Short Answer

Expert verified

The skeletal structure is given as,

Step by step solution

01

Drawing of structures from IUPAC name

We can draw the structures from the IUPAC name of the compound by using the following rules and this is as follows.

a. Find out the longest chain and do the numbering.

Ex, If a compound is 3-methyl-butan-1-ol, that means the longest chain name is butane.

b. Find out the positions of substituting groups in the longest chain.

Ex- for the above compound, Methyl is attached to the 3rd position and the hydroxyl group is in 1 position of the skeletal chain or the main chain, So, I should be written in the 3rd position and hydroxyl group in 1st position.

c. 鈥渘鈥� in a compound designated non-branched compound.

d. N, N means the substitution of groups to N of amine groups.

02

 Step 2: Skeletal structure

The skeletal structure is given as follows,

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Most popular questions from this chapter

Draw curved arrows to show the movement of the electrons in each step of the following reaction sequences. (Hint: You can tell how to draw the arrows for each step by looking at the products that are formed in that step as a result of the movement of electrons.)

From the following rate constants, determined at five temperatures, calculate the experimental energy of activation and 螖G卤,螖贬卤,补苍诲螖厂卤for the reaction at 30掳C:

a. For a reaction with 鈭咹掳 = -12 kcal/mol and 鈭哠掳 = 0.01kcal mol-1 K-1, calculate the 鈭咷掳 and the equilibrium constant at: 1. 30 掳C and 2. 150 掳C.

b. How does 鈭咷掳 change as T increases?

c. How does Keq change as T increases?

Which bond is stronger? Briefly explain why.

In a reaction in which reactant A is in equilibrium with product B at

25 掳C, what relative amounts of A and B are present at equilibrium if 螖G掳 at 25 掳C is

a. 2.72 kcal/mol?

b. 0.65 kcal/mol?

c. -2.72 kcal/mol?

d. -0.65 kcal/mol?
See all solutions

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