CPA vs CCA attack of an encryption scheme Semantic Security is a concept that made New Cryptography, (that is Cryptography invented after 1975) into a rigorous science. It connected old cryptography (that is Shannon's information theory-based cryptography) to New Cryptography. 

Suppose that we use a randomly chosen 128-bit IV  to encipher a 128-bit message $m$, and AES with a randomly chosen key $k$, to generate a 256-bit cipher-text computed as: $IV \| \mathrm{AES}(k,(IV \oplus m))$. Here $\|$ means concatenation and $\oplus$ means exclusive or.

Why is this scheme semantically secure against an eavesdropper, but not  against an active attacker? -- i.e why is it secure against Chosen Plaintext Attack, but not against Chosen Ciphertext Attack?
To answer the question, its necessary to invoke the following security definition, which requires that for any encryption scheme $\pi$ to have indistinguishable encryptions under chosen ciphertext attack (viz., for the scheme to be CCA secure), the following condition, which I'll describe, must hold in the so called "CCA distinguishing game" (throughout, I'm using the notation of Katz and Lindell, and the CCA property is more expansively described in their text Intro to Modern Cryptography, 2nd ed. CRC Press):

$\textbf{Pr}[\textbf{PrivK}_{\mathcal{A},\pi}^{cca}(n) = 1] \leq 1/2 +  \text{negl}(n)$ 

Here $\textbf{Pr}[E=e]$ denotes the probability that experiment $E$ has outcome $e$, and $\textbf{PrivK}_{\mathcal{A},\pi}^{cca}(n) = 1$ denotes the outcome where the CCA adversary $\mathcal{A}$ wins the CCA distinguising game ($K$ is an encryption key of length $n$ and $\text{negl}(n)$ is a "negligible'' function in $n$). Also, throughout we assume $\mathcal{A}$ is PPT. 

(this definition needs to be expanded to allow for encrpytions and decryptions of multiple messages in an analogous vein to the "Left-Right" or LR-Oracle CPA security game described here - https://cseweb.ucsd.edu/~mihir/cse207/w-se.pdf, but the above definition is sufficient for illustration purposes)

So to restate, the CCA security definition requires that the above condition for the scheme $\pi$ must hold in the CCA distinguishing game $\textbf{PrivK}_{\mathcal{A},\pi}^{cca}(n)$ in order for the scheme to be CCA secure. The rules of the game are that the adversary $\mathcal{A}$  has oracle access to the encryption and decryption functions, and it proceeds by her selecting 2 equal length messages $m_0$ and $m_1$  and submitting them to the encryption oracle, which flips a coin and based on the outcome sets a bit $b$ to either 0 or 1. Based on the value of $b$, the oracle encrypts either $m_0$ or $m_1$ and returns the resulting ciphertext $c$ to $\mathcal{A}$ (which is also called the challenge ciphertext). $\mathcal{A}$ can continue to invoke the encryption and decryption oracle on any messages other than the challenge ciphertext. At the end, $\mathcal{A}$ outputs a bit $b^{\prime}$. The adversary wins the game, which is denoted by the condition $\textbf{PrivK}_{\mathcal{A},\pi}^{cca}(n) = 1$ when $b^{\prime} = b$. 

Now consider a CPA secure counter mode encryption scheme such as AES-128-CTR. $\mathcal{A}$ can choose messages $m_0 = 0^{128}$ and $m_1 = 1^{128}$ and submit them to the encryption oracle. Suppose that the oracle sets $b = 0$ and returns the challenge ciphertext $c$. The adversary can then flip the first bit of c resulting in a ciphertext $c^{\prime}$. Since $ c^{\prime} \ne c$, $\mathcal{A}$ can legitimately submit this to the decryption oracle which returns the plaintext $10^{127}$, at which point $\mathcal{A}$ outputs $b^{\prime} = 0$ and can trivially win the game every time thereby violating the CCA security condition $\textbf{Pr}[\textbf{PrivK}_{\mathcal{A},\pi}^{cca}(n) = 1] \leq 1/2 +  \text{negl}(n)$