(a) Give the empirical formula and structural formula for a cyclic ether containing four carbon atoms in the ring. (b) Write the structural formula for a straight-chain compound that is a structural isomer of your answer to part (a).

The cyclic ether will have four carbon atoms in the ring and an oxygen atom to complete the ring (as ethers have an oxygen atom between two carbon atoms). The empirical formula for this cyclic ether can be found by counting the number of elements in the molecule and then simplifying the ratio. We have: - Four carbon atoms (C) - One oxygen atom (O) - Eight hydrogen atoms (H) attached to the four carbon atoms in the ring (since each carbon atom in a cyclic ether has two hydrogen atoms) This gives the empirical formula of \(C_4H_8O\). For the structural formula of the cyclic ether, 1. Start with 4 carbon atoms arranged in a ring. 2. Connect an oxygen atom to complete the ring. 3. Attach two hydrogen atoms to each carbon atom. The structural formula can be drawn as: ``` H H H H | | | | C-C-O-C-C | | | | H H H H ```

An isomer of the cyclic ether will have the same empirical formula, but a different structural arrangement of its atoms. To create a straight-chain isomer, we can break the four-carbon cyclic ether and convert it into a straight chain with an oxygen atom in between two carbon groups. Here's how to create the structural formula for a straight-chain isomer: 1. Start with 3 carbon atoms in a straight chain, the first and second carbon atoms connected to each other. 2. Connect an oxygen atom to the second carbon atom. 3. Connect a fourth carbon atom to the oxygen atom. 4. Attach the hydrogen atoms to the carbon atoms according to the valency (two hydrogen atoms for the first and last carbon and one hydrogen atom for the middle carbon). The structural formula can be drawn as: ``` H H H | | | H C-C-C-O-C | | | H H H H ``` So, this is the structural formula of a straight-chain compound that is a structural isomer of the cyclic ether.